# Why is group theory axiomatized with operations?

The notion of 'operation' seems so arbitrary. In textbooks on group theory, and for example on Wikipedia, they define the group multiplication operation $\circ$ as having properties of:

1. Closure: For all $a, b\in G$, the result of the operation, $a\circ b$, is also in $G$.

2. Associativity: For all $a, b, c\in G$, $(a\circ b)\circ c = a\circ (b\circ c)$.

3. Identity element: There exists an element $e \in G$ such that, for every element $a \in G$, the equation $e\circ a = a\circ e = a$ holds.

4. Inverse element: For each $a \in G$, there exists an element $b \in G$, such that $a \circ b = b \circ a = e$, where e is the identity element.

I do not understand why axiom (1) is used.

Often, axiom (1) is replaced by defining a function $$\circ:G\times G \rightarrow G$$ and stating that axioms (2)-(4) hold for this binary function.

Here the closure axiom becomes unnecessary, but more importantly, it is more clear what an operation is, and why the properties of associativity and commutativity are actually special. Furthermore, students will be more careful when doing proofs, because they know that the properties they are using are not valid for ordinary functions.

I'm not questioning the correctness of group theory, but this is something I've wondered about for a while.

• Definitions of group do not usually talk about "closure". – Lord Shark the Unknown Feb 17 '18 at 18:25
• I think the primary reason is the belief - correct or not - that speaking in terms of "binary operations" is clearer for students with no prior experience than jumping right into maps $G\times G\rightarrow G$. The point is that "binary operation" is a phrase that they've (probably) heard before, and may be able to absorb more quickly than general function notation. I'm not sure whether this is right, but I think that's likely a motivation (there may also be historical precedent, I'm not sure how the original definition of an abstract group was phrased). – Noah Schweber Feb 17 '18 at 18:28
• You said "bla bla bla", but what are the axioms that you're not understanding? – TPace Feb 17 '18 at 18:31
• This may help: math.stackexchange.com/questions/2377598/… – Ethan Bolker Feb 17 '18 at 18:41
• The OP is not "tilting at a straw man". Sure, the 3 books you found don't use this definition. However, Wikipedia, ProofWiki, WolframMathWorld, the Open University, Queen Mary University of London, and Berkeley all do. – user1729 Feb 18 '18 at 10:16

## 2 Answers

These closure properties are explicitly stated because it is important for subgroups. Indeed, if $H$ is a subgroup of $G,\times$ one needs that $h\times h'\in H$. In this sense the operation is closed. In other words, one needs that $\text{im}(\times_{\mid H\times H})\subseteq H$. It easier to say that $H$ is closed.

Perhaps one reason is the "binary operation" definition is completely abstract, however it is helpful to be concrete in a first course in group theory. For example, suppose I want to show that all real-valued invertible $n\times n$ matrices, denoted $GL_n(\mathbb{R})\subset M_n(\mathbb{R})$, form a group. Then what I have to do is think about the function $$\circ: GL_n(\mathbb{R})\times GL_n(\mathbb{R})\rightarrow M_n(\mathbb{R})$$ and show that the image of this function is precisely $GL_n(\mathbb{R})$; that is, show closure. Here, $M_n(\mathbb{R})$ is not a group so we are not talking about subgroups of groups but something slightly more complicated.